Prove that if $a,b,$ and $c$ are positive real numbers, then $\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a} \geq ab + bc + ca$. 
Prove that if $a,b,$ and $c$ are positive real numbers, then $\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a} \geq ab + bc + ca$.

I tried AM-GM and it doesn't look like AM-GM or Cauchy-Schwarz work here. The $ab+bc+ca$ reminds of a cyclic expression, so that may help by factoring the inequality and getting a true statement. 
 A: First I'll prove a Lemma: $a^2+b^2+c^2\ge ab+bc+ca$ for all $a,b,c\in\mathbb R$.
Proof: it follows from the Rearrangement Inequality, because $(a,b,c)$ and $(a,b,c)$ are similarly sorted.
Or notice that it's equivalent to $\frac{1}{2}\left((a-b)^2+(b-c)^2+(c-a)^2\right)\ge 0$, which is true; or add the following inequalities: $a^2+b^2\ge 2ab$, $b^2+c^2\ge 2bc$, $c^2+a^2\ge 2ca$. $\ \square$
Your inequality is cyclic. Wlog there are two cases:

*

*$a\ge b\ge c$. Then $(a,b,c)$ and $(1/a,1/b,1/c)$ are oppositely sorted, so
$$\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge \frac{a^3}{a}+\frac{b^3}{b}+\frac{c^3}{c}=a^2+b^2+c^2$$
Now use the Lemma.


*$a\ge c\ge b$. Then $(a,c,b)$ and $(1/a,1/c,1/b)$ are oppositely sorted, so  $$\frac{a^3}{b}+\frac{c^3}{a}+\frac{b^3}{c}\ge \frac{a^3}{a}+\frac{c^3}{c}+\frac{b^3}{b}=a^2+c^2+b^2$$
Now use the Lemma.
In your inequality, equality holds if and only if $a=b=c$.

Another proof: By Hölder's inequality:
$$\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge \frac{(a+b+c)^3}{3(a+b+c)}=\frac{(a+b+c)^2}{3}$$
Also $(a+b+c)^2\ge 3(ab+bc+ca)$, because this is equivalent to $a^2+b^2+c^2\ge ab+bc+ca$ (see the Lemma).
A: It is actually very simple. Use nothing but AM-GM.
$$\frac{a^3}{b} + ab \geq 2a^2$$
$$\frac{b^3}{c} + bc \geq 2b^2$$
$$\frac{c^3}{a} + ac \geq 2c^2$$
$$LHS + (ab+bc+ac) \geq 2(a^2+b^2+c^2) \geq 2(ab + bc +ac)$$
We are done.
A: For positive $x$, $y$, $z$, $a$, $b$, and $c$, note that that 
\begin{align*}
(x+y+z)^2 &= \left(\frac{x}{\sqrt{a}}\sqrt{a} + \frac{y}{\sqrt{b}}\sqrt{b} +\frac{z}{\sqrt{c}}\sqrt{c}\right)^2\\
&\le  \left(\frac{x^2}{a} + \frac{y^2}{b} +\frac{z^2}{c}\right)(a+b+c).
\end{align*}
That is,
\begin{align*}
\frac{x^2}{a} + \frac{y^2}{b} +\frac{z^2}{c} \ge \frac{(x+y+z)^2}{a+b+c}.
\end{align*}
Then
\begin{align*}
\frac{a^3}{b} + \frac{b^3}{c} +\frac{c^3}{a} &=\frac{a^4}{ab} + \frac{b^4}{bc} +\frac{c^4}{ac}\\
&\ge \frac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ac}\\
&\ge \frac{\left(ab+bc+ac\right)^2}{ab+bc+ac}\\
&=ab+bc+ac.
\end{align*}
A: The statement is false if you take $a = b = c = 1$.
A: your inequalitiy is equivalent to
$$a^4c+b^4a+c^4b-a^2b^2c-a^2bc^2-ab^2c^2\geq 0$$
this is equivalent to
$$(a^2-b^2)(a^2c-bc^2)+(b^2-c^2)(ab^2-bc^2)\geq 0$$ we assume that $$a\geq b\geq c$$ thus $$a^2\geq b^2$$ and $$a^2\geq bc$$ and $$b^2\geq c^2$$ and $$ab\geq c^2$$
in the case $$a\geq c\geq b$$ we have
$$(a^2-c^2)(a^2c-c^2b)+(a^2c-b^2a)(c^2-b^2)\geq 0$$
A: $$
\begin{align}
&\left(\frac{a^3}b+\frac{b^3}c+\frac{c^3}a\right)-(ab+bc+ca)\\
&=\frac ab\left(a^2-b^2\right)+\frac bc\left(b^2-c^2\right)+\frac ca\left(c^2-a^2\right)\\[3pt]
&=\left(\frac ab-1\right)\left(a^2-b^2\right)+\left(\frac bc-1\right)\left(b^2-c^2\right)+\left(\frac ca-1\right)\left(c^2-a^2\right)\\
&=\frac{(a-b)^2(a+b)}b+\frac{(b-c)^2(b+c)}c+\frac{(c-a)^2(c+a)}a\\[6pt]
&\ge0
\end{align}
$$
This also shows that equality requires $a=b=c$.
A: We have:
$$ \frac{a^4}{ab}+\frac{b^4}{bc}+\frac{c^4}{ca} ≥ ab + bc + ca $$
Also note that
$$ \frac{a^4}{ab}+\frac{b^4}{bc}+\frac{c^4}{ca} ≥ \frac{(a^2+b^2+c^2)^2}{ab+bc+ca}$$
(Because of the Cauchy Schwarz lemma or the Titu's Lemma)
So if we can prove that 
$$\frac{(a^2+b^2+c^2)^2}{ab+bc+ca} ≥ ab + bc + ca$$
Then our work is done; so
$$ (a^2+b^2+c^2)^2 ≥ (ab+bc+ca)^2$$
$$ ⇒ a^2+b^2+c^2 ≥ ab+bc+ca   \ \square$$
This can simply be proven using the rearrangement inequality.
